Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100;
vector<int> G[N], pre[N];
int posi[N], depth[N], fa[N][18], n;
vector<int> st[N][18];
void get(vector<int>& a, vector<int>& b) {
  for (auto v : a) b.push_back(v);
  sort(b.begin(), b.end());
  while (b.size() > 10) b.pop_back();
}
void dfs(int u, int f, int dep) {
  depth[u] = dep;
  fa[u][0] = f;
  if (f != -1) {
    get(pre[f], st[u][0]);
  }
  for (auto v : G[u])
    if (v != f) dfs(v, u, dep + 1);
}
void build() {
  for (int i = 0; i < 18 - 1; i++)
    for (int j = 1; j <= n; j++) {
      fa[j][i + 1] = (fa[j][i] == -1 ? -1 : fa[fa[j][i]][i]);
      if (fa[j][i + 1] != -1) {
        get(st[j][i], st[j][i + 1]);
        get(st[fa[j][i]][i], st[j][i + 1]);
      }
    }
}
void show(int a, vector<int>& ans) {
  int lim = min(a, (int)ans.size());
  printf("%d", lim);
  for (int i = 0; i < (int)lim; i++) printf(" %d", ans[i]);
  printf("\n");
}
int cal(int u, int v, int a) {
  if (depth[u] > depth[v]) swap(u, v);
  int cha = depth[v] - depth[u];
  vector<int> ans;
  get(pre[v], ans);
  for (int i = 18 - 1; i >= 0; i--) {
    if ((cha >> i) & 1) {
      get(st[v][i], ans);
      v = fa[v][i];
    }
  }
  if (u == v) {
    show(a, ans);
    return u;
  }
  get(pre[u], ans);
  for (int i = 18 - 1; i >= 0; i--) {
    if (fa[v][i] != fa[u][i]) {
      get(st[v][i], ans);
      get(st[u][i], ans);
      v = fa[v][i];
      u = fa[u][i];
    }
  }
  get(st[u][0], ans);
  show(a, ans);
  return fa[u][0];
}
int m, q;
void read() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < (int)n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= (int)m; i++)
    scanf("%d", &posi[i]), pre[posi[i]].push_back(i);
  dfs(1, -1, 0);
  build();
  for (int i = 1; i <= (int)n; i++)
    if (pre[i].size()) {
      sort(pre[i].begin(), pre[i].end());
      while (pre[i].size() > 10) pre[i].pop_back();
    }
  int u, v, a;
  while (q--) {
    scanf("%d %d %d", &u, &v, &a);
    cal(u, v, a);
  }
}
int main() {
  read();
  return 0;
}